poj 1017 Packets【贪心】

最新推荐文章于 2022-11-28 14:14:36 发布

原创最新推荐文章于 2022-11-28 14:14:36 发布 · 637 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#input #output #file #each #null

poj 专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一个工厂面临的问题：如何将不同大小的产品最有效地装入统一规格的大包裹中以减少运输成本。通过一个C++程序实现了寻找最小数量包裹的解决方案，特别关注了如何处理2*2规格产品的装箱难题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Description

A factory produces products packed in square packets of the same height h and of the sizes 1*1, 2*2, 3*3, 4*4, 5*5, 6*6. These products are always delivered to customers in the square parcels of the same height h as the products have and of the size 6*6. Because of the expenses it is the interest of the factory as well as of the customer to minimize the number of parcels necessary to deliver the ordered products from the factory to the customer. A good program solving the problem of finding the minimal number of parcels necessary to deliver the given products according to an order would save a lot of money. You are asked to make such a program.

Input

The input file consists of several lines specifying orders. Each line specifies one order. Orders are described by six integers separated by one space representing successively the number of packets of individual size from the smallest size 1*1 to the biggest size 6*6. The end of the input file is indicated by the line containing six zeros.

Output

The output file contains one line for each line in the input file. This line contains the minimal number of parcels into which the order from the corresponding line of the input file can be packed. There is no line in the output file corresponding to the last ``null'' line of the input file.

Sample Input

0 0 4 0 0 1 
7 5 1 0 0 0 
0 0 0 0 0 0

Sample Output

2 
1

代码如下：

#include<iostream> using namespace std; int pk[7]; int pk3[4]={0,5,3,1}; //记录被3*3的占用之后，还剩多少给2*2的使用 int main() { int i; bool flag; for(i=1;i<=6;i++) { scanf("%d",&pk[i]); if(pk[i]) flag=true; } while(flag) { int sum=0; flag=false; sum=pk[4]+pk[5]+pk[6]+(pk[3]+3)/4;//向上调整 int left2=pk[4]*5+pk3[pk[3]%4]; //还可以装多少个2*2的 if(left2<pk[2]) sum+=(pk[2]-left2+8)/9;//向上调整 int left1=sum*36-pk[2]*4-pk[3]*9-pk[4]*16-pk[5]*25-pk[6]*36; //还可以装多少个1*1的，妙啊！ if(left1<pk[1]) sum+=(pk[1]-left1+35)/36;//向上调整 printf("%d/n",sum); for(i=1;i<=6;i++) { scanf("%d",&pk[i]); if(pk[i]) flag=true; } } return 0; }

一切尽在代码中。

其实这道题的关键就是要解决2*2规格的。处理的时候要注意向上调整。

虽然如此我还是错了很多次，然后找了一篇牛人的看了才发现自己的一些小错误。真是一道不水的水题。